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Differentiation II

Implicit differentiation

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Summary

Implicit differentiation

In a nutshell

If a function cannot easily be arranged such that you have it in the form $y=f(x)$ or $x=g(y)$ , then you cannot differentiate explicitly, which is what you have been doing so far. Instead, you can differentiate the function as it is implicitly.

Explicit vs. implicit

A function of the form $y=f(x)$ or $x=g(y)$ is given explicitly. On one side is a $y$ or an $x$ , and on the other side is a function in terms of the other variable. A function in the form $f(x,y)$ is a function in terms of both $x$ and $y$ . The equation $0=f(x,y)$ is given implicitly.

Example 1

The following functions are given implicitly.

$4xy+x^2-y^3=6\\\sin(x-y)=0\\\frac{x}{y^2}-6xy=-9^x$ 

Using the chain rule

Recall that when you denote differentiation, you use $\frac{\text d}{\text dx}$ , meaning to differentiate with respect to $x$ . It is commonly seem as $\frac{\text dy}{\text dx}$ . meaning you are differentiating $y$ with respect to $x$ .

Suppose you have a function of $y$ , $f(y)$ , but since it is part of a larger, implicit function, you need to differentiate with respect to $x$ , you will use the chain rule. In this case, you will use it in this form:

$\frac{\text d}{\text dx}\rightarrow\frac{\text dy}{\text dx}\times\frac{\text d}{\text dy}$

In other words, you have converted the differentiation operation into one in terms of $y$ . All you have to do is to multiply by $\frac{\text dy}{\text dx}$ . The general form when differentiating a function $f(y)$ with respect to $x$ is

$\boxed{\frac{\text d}{\text dx}(f(y))=f'(y)\frac{\text dy}{\text dx}}$

Example 2

Differentiate the following with respect to $x$ :

$5y^2+7y=4x^3$ 

Apply $\frac{\text d}{\text dx}$ to both sides:

$\frac{\text d}{\text dx}(5y^2+7y)=\frac{\text d}{\text dx}(4x^3)$ 

The right-hand side can be done as expected, since it is in terms of $x$ :

$\frac{\text d}{\text dx}(5y^2+7y)=12x^2$

The left-hand side however requires the chain rule:

$\begin{aligned}\frac{\text dy}{\text dx}\times\frac{\text d}{\text dy}(5y^2+7y)&=12x^2\\\frac{\text dy}{\text dx}(10y+7)&=12x^2\end{aligned}$ 

Rearranging this to obtain the equation for the gradient gives:

$\underline{\frac{\text dy}{\text dx}=\frac{12x^2}{10y+7}}$ 

Note: When implicitly differentiating, it is common that your expression for $\frac{\text dy}{\text dx}$ will be a function of both $x$ and $y$ .

Example 3

Differentiate the following with respect to $x$ :

$0=2y+xy^3$ 

Apply $\frac{\text d}{\text dx}$ to both sides:

$\begin{aligned}\frac{\text d}{\text dx}(0)&=\frac{\text d}{\text dx}(2y+xy^3)\\0&=\frac{\text d}{\text dx}(2y+xy^3)\end{aligned}$ 

To differentiate the right-hand side, approach term by term:

$\begin{aligned}0&=\frac{\text d}{\text dx}(2y)+\frac{\text d}{\text dx}(xy^3)\\&=\frac{\text dy}{\text dx}\frac{\text d}{\text dy}(2y)+\frac{\text d}{\text dx}(xy^3)\\&=2\frac{\text dy}{\text dx}+\frac{\text d}{\text dx}(xy^3)\end{aligned}$

To differentiate the second term, you will use the product rule:

$\begin{aligned}0&=2\frac{\text dy}{\text dx}+\frac{\text d}{\text dx}(xy^3)\\&=2\frac{\text dy}{\text dx}+\left(\frac{\text d}{\text dx}(x)\right)y^3+x\left(\frac{\text d}{\text dx}(y^3)\right)\\&=2\frac{\text dy}{\text dx}+(1)y^3+x\left(\frac{\text dy}{\text dx}\frac{\text d}{\text dy}(y^3)\right)\\&=2\frac{\text dy}{\text dx}+y^3+x\left(\frac{\text dy}{\text dx}(3y^2)\right)\\&=2\frac{\text dy}{\text dx}+y^3+3xy^2\frac{\text dy}{\text dx}\end{aligned}$

You can now rearrange this to give an equation for $\frac{\text dy}{\text dx}$ :

$\begin{aligned}0&=2\frac{\text dy}{\text dx}+y^3+3xy^2\frac{\text dy}{\text dx}\\&=\frac{\text dy}{\text dx}(2+3xy^2)+y^3\\\frac{\text dy}{\text dx}&=\underline{-\frac{y^3}{2+3xy^2}}\end{aligned}$

Common derivatives

Below are some frequently occuring derivatives. You may want to commit them to memory.

$\begin{aligned}\frac{\text d}{\text dx}(y^n)&=ny^{n-1}\frac{\text dy}{\text dx}\\\frac{\text d}{\text dx}(xy)&=x\frac{\text dy}{\text dx}+y\end{aligned}$

where $n$ is a constant.

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Differentiating functions with multiple terms

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